Now why doesn't the entropy change? Well remember ##S = k_B \ln \Omega(\bar{E})## and ##\Omega(\bar{E})## is the volume between ##\bar{E}## and ##\bar{E} + \delta\bar{E}## in phase space i.e. of an infinitesimal energy shell of microstates accessible to the system. So if you shift all energy eigenvalues of the system by the same constant, thus shifting the average energy by the same amount, then the volume in phase space in between two infinitesimally separated values of average energy will remain unchanged.

To give another view to WannabeNewton's very good explanation, one could say that a shift in energy levels of a system will not alter the occupation of the levels and therefore not change the entropy of the system. I guess your question seems like a specific case of the consequence that a perturbation to a system in the form of heat changes the distribution/occupation of levels without changing the energy levels themselves, whereas a perturbation in the form of reversible work changes the energy levels without affecting their distribution.
In other words the changes in energy levels of a system must come from a change to the Hamiltonian i.e. change in interactions between the particles or interaction with external fields, so from the effect of work on a system, and this doesn't change entropy. A change in entropy arises from the effect of heat and this changes the energy of the system by repopulating to higher energy levels without changing the levels themselves (doesn't change the form of the Hamiltonian).